Question

A horizontal bridge is built across a river. A student standing on the bridge throws a small ball vertically upwards with a velocity \(4\ ms^{-1}\). The ball strikes the water surface after \(4s\). The height of bridge above water surface is
(Take: \(g=10\ ms^{-2}\))

• A. 60 m
• B. 64 m
• C. 68 m
• D. 56 m

Answer 1

The Correct Option is B

To find the height of the bridge, we can use the kinematic equation for motion under uniform acceleration: \(s = ut + \frac{1}{2}at^2\).

Given:

• Initial velocity, \(u = 4\ ms^{-1}\) (upward)
• Time, \(t = 4\ s\)
• Acceleration due to gravity, \(a = -10\ ms^{-2}\) (since the direction is opposite to the initial velocity)

Here, the displacement \(s\) will be negative as the ball goes below the bridge after it is thrown upwards. Thus, the equation becomes:

\(s = 4\cdot4 + \frac{1}{2}\cdot(-10)\cdot4^2\)

\(s = 16 - 80\)

\(s = -64\ m\)

Since the displacement is negative, this indicates that the ball's final position is 64 m below the starting position. Thus, the height of the bridge is 64 m.

The correct answer is 64 m.

Answer 2

Given: 

• A small ball is thrown vertically upwards with a velocity of 4 m/s from a horizontal bridge.
• The ball strikes the water surface after 4 seconds.
• Acceleration due to gravity,
  \(g = 10 m/s^2\)
  \(u = 4m/s\)
  \(g = -10 m/s^2\)

Let's assume the initial height of the ball above the water surface to be h.

We know that the distance traveled by the ball in the upward direction is equal to that traveled by the ball in the downward direction.
t = 4 seconds

Displacement= -ve, and let this displacement be =S

\(S = Ut + \frac{1}{2} \times a(t^2)\)
\(-S = 4 . 4 + \frac{1}{2} \times (-10) \times16\)
\(-S = 16 - 16(5)\)
\(-S = -64\)
\(S = 64 m\)

Hence The height of the bridge above the water surface is 64 meters.

Answer 3

We know that:
\(S=ut+\frac{1}{2}at^2\)
\(-H=4\times4-\frac{1}{2}\times10\times4^2\)
\(-H=16-80\)
\(H=64\,m\)

So, the correct option is (B): \(60 m\)